Question

A projectile is projected with velocity v at an angle theta with horizontal find the velocity of projectile at any instant t

Answer 1

$a \: projectile \: is \: thrown \: with \: velocity \\ \: v \: at \: angle \: \alpha \\ we \: know \: that \\ v = \sqrt{( {horizontal \: component) }^{2} + {vertical}^{2} }$
$now \: its \: horizontal \: velocity \: will \: remain \\ constant \\ \: throught \: out \: the \: motion \\ \: now \: let \: horizontal \: componet \: is \: vcos \alpha \\ and \: vertical \: component \: is \: vsin \alpha \\ vcos \alpha \: will \: remain \: same \: as \: there \: is \: no \\ force \: in \: \\ horizontal \: direction \\ vsin \alpha \: will \: decrease \: upto \: max \: height \: and \ \\ then \: i t\\ will \: increase \: \\ so \: vsin \alpha \: - gt \: is \: vertical \: component \: \\ of \: velocity \: at \: \: any \: time \: t \\ so \: velocity \: at \: any \: time \: t \\ = \sqrt{ {(vcos \alpha) }^{2} + {(vsin \alpha - gt)}^{2} }$